Is there a sequence $\{f_n\}$ holomorphic on D(0,2) such that $f_n\to \bar{z}^3$ uniformly on $\{|z|=1\}\cup\{|z|=\frac{1}{2}\}$. I notice that on $\{|z|=1\}$, $\bar{z}^3=\frac{1}{z^3}$, $\{|z|=\frac{1}{2}\}$, $\bar{z}^3=\frac{1}{64z^3}$, seemingly antiholomorphic function is actually holomorphic on the 2 circles. Runge's theorem requires connectedness on complement and hence not applicable.
Thank you!
No. If such a sequence existed, define $g_n(z)=z^2f_n(z)$. Then, $(g_n)_{n\in\mathbb N}$ converges uniformly to $\frac1z$ on $\{z\in\mathbb{C}\,|\,\lvert z\rvert=1\}$. Now, consider the loop $\gamma\colon[0,2\pi]\longrightarrow\mathbb C$ defined by $\gamma(t)=e^{it}$. Then$$(\forall n\in\mathbb{N}):\int_\gamma g_n(z)\,\mathrm dz=0,$$but $\int_\gamma\frac1z\,\mathrm dz=2\pi i$.